Additional Notes: Differential Equations | Engineering Mathematics for Electrical Engineering - Electrical Engineering (EE) PDF Download

Cauchy-Euler Equations

Another class of solvable linear differential equations that is of interest are the Cauchy-Euler type of equations, also referred to in some books as Euler's equation.
These are given by:
ax²y''(x) + bxy'(x) + cy(x) = 0     .......(2.5.1)
Note that in such equations, the power of x in each of the coefficients matches the order of the derivative in that term. These equations are solved in a manner similar to the constant coefficient equations.
One begins by making the guess y(x) = xʳ. Inserting this function and its derivatives,
y'(x) = r xʳ⁻¹,
y''(x) = r(r - 1) xʳ⁻² into Equation 2.5.1, we have:
[ar(r - 1) + br + c] xʳ = 0
Since this has to be true for all x in the problem domain, we obtain the characteristic equation:
ar(r - 1) + br + c = 0
The solutions of Cauchy-Euler equations can be found using this characteristic equation. Just like the constant coefficient differential equation, we have a quadratic equation, and the nature of the roots again leads to three classes of solutions. If there are two real, distinct roots, then the general solution takes the form:
y(x) = c₁ xʳ¹ + c₂ xʳ²

Example1: Find the general solution:
x²y'' + 5xy' + 12y = 0

Solution: As with the constant coefficient equations, we begin by writing down the characteristic equation. Doing a simple computation,

0 = r(r - 1) + 5r + 12
= r² + 4r + 12
= (r + 2)² + 8
-8 = (r + 2)²

One determines the roots as r = -2 ± 2√2i. Therefore, the general solution is:

y(x) = [c₁ cos(2√2 ln |x|) + c₂ sin(2√2 ln |x|)] x⁻²

Deriving the solution for Case 2 for the Cauchy-Euler equations works in the same way as the second case for constant coefficient equations, but it is a bit messier.

First, note that for the real root r = r₁, the characteristic equation has to factor as (r - r₁)² = 0. Expanding, we have:

r² - 2r₁r + r₁² = 0

The general characteristic equation is:

ar(r - 1) + br + c = 0

Dividing this equation by a and rewriting, we have:

r² + (b/a - 1) r + c/a = 0

Comparing equations, we find:

b/a = 1 - 2r₁,
c/a = r₁²

So, the Cauchy-Euler equation for this case can be written in the form:

x²y'' + (1 - 2r₁) xy' + r₁² y = 0

Now we seek the second linearly independent solution in the form y₂(x) = v(x) xʳ.
We first list this function and its derivatives:

y₂(x) = v xʳ
y₂'(x) = (xʳ v' + r₁ v) xʳ⁻¹
y₂''(x) = (xʳ v'' + 2r₁ xʳ v' + r₁ (r₁ - 1) v) xʳ⁻²   (Equation 2.5.2)

Inserting these forms into the differential equation, we have:

0 = x² y'' + (1 - 2r₁) x y' + r₁² y
= (xʳ v'' + 0) xʳ⁺¹

Thus, we need to solve the equation:

x v'' + v' = 0

Or,

v'/v' = -1/x

Integrating, we have:

ln|v| = -ln|x| + C

where A = ±e^C absorbs C and the signs from the absolute values. Exponentiating, we obtain one last differential equation to solve:

v' = A / x

Thus,

v(x) = A ln|x| + k

So, we have found that the second linearly independent equation can be written as:

y₂(x) = xʳ ln|x|.

Therefore, the general solution is found as:

y(x) = (c₁ + c₂ ln|x|) xʳ.

For one root, r₁ = r₂ = r, the general solution is of the form:

y(x) = (c₁ + c₂ ln|x|) xʳ.

Example 2: Solve the initial value problem: t²y'' + 3ty' + y = 0, with initial conditions y(1) = 0, y'(1) = 1.

Solution: The characteristic equation is:
r(r - 1) + 3r + 1 = 0
or
r² + 2r + 1 = 0

Since there is only one real root, r = -1, the general solution is:
y(t) = (c₁ + c₂ ln |t|) t⁻¹

Using the initial condition y(1) = 0, we find c₁ = 0.
Using y'(1) = 1, we find c₂ = 1.

Thus, the solution is y(t) = ln |t| t⁻¹.

Now, for complex conjugate roots r = α ± iβ, the general solution takes the form:
y(x) = x^α [c₁ cos(β ln |x|) + c₂ sin(β ln |x|)]

Example 3: Solve: x²y'' - xy' + 5y = 0.

Solution: The characteristic equation is:
r(r - 1) - r + 5 = 0
or
r² - 2r + 5 = 0

The roots are r₁,₂ = 1 ± 2i.
Thus, the general solution is:
y(x) = x¹ [c₁ cos(2 ln |x|) + c₂ sin(2 ln |x|)]

Classification of Roots of the Characteristic Equation for Cauchy-Euler Differential Equations

1. Real, distinct roots r₁, r₂.
   In this case, the solutions corresponding to each root are linearly independent. Therefore, the general solution is simply:
   y(x) = c₁ xʳ¹ + c₂ xʳ²

2. Real, equal roots r₁ = r₂ = r.
   In this case, the solutions corresponding to each root are linearly dependent. To find a second linearly independent solution, one uses the Method of Reduction of Order. This gives the second solution as xʳ ln |x|. Therefore, the general solution is found as:
   y(x) = (c₁ + c₂ ln |x|) xʳ

3. Complex conjugate roots r₁, r₂ = α ± iβ.
   In this case, the solutions corresponding to each root are linearly independent. These complex exponentials can be rewritten in terms of trigonometric functions. Namely, one has that xᵅ cos(β ln |x|) and xᵅ sin(β ln |x|) are two linearly independent solutions. Therefore, the general solution becomes:
   y(x) = xᵅ [c₁ cos(β ln |x|) + c₂ sin(β ln |x|)]

Nonhomogeneous Cauchy-Euler Equations

We can also solve some nonhomogeneous Cauchy-Euler equations using the Method of Undetermined Coefficients or the Method of Variation of Parameters. We will demonstrate this with a couple of examples.

Example 1: Find the solution of x²y'' - xy' - 3y = 2x².

Solution: First, we find the solution of the homogeneous equation. The characteristic equation is:
r² - 2r - 3 = 0
The roots are r = -1, 3, so the solution is:
yₕ(x) = c₁ x⁻¹ + c₂ x³

Now, we need a particular solution. Let's guess yₚ(x) = Ax². Inserting this guess into the nonhomogeneous differential equation:

2x² = 2x²y'' - xy' - 3y = 2x²
= 2A x² - 2A x² - 3A x²
= -3A x²

Thus, A = -2/3.
Therefore, the general solution of the problem is:
y(x) = c₁ x⁻¹ + c₂ x³ - (2/3) x²

Example 2: Find the solution of x²y'' - xy' - 3y = 2x³.

Solution: In this case, the nonhomogeneous term is a solution of the homogeneous equation, which we solved in the last example. So, we will need a modification of the method.

We have a problem of the form:
ax²y'' + bxy' + cy = dxʳ

where r is a solution of ar(r - 1) + br + c = 0.
Let’s guess a solution of the form y = Axʳ ln x.

Then, substituting in the equation, it reduces to:
Azʳ (2ar - a + b) = dzʳ. (You should verify this for yourself.)

With this in mind, solving the problem:
Let yₚ = Ax³ ln x. Inserting into the equation:
4Ax³ = 2x³
Thus, A = 1/2.

The general solution of the problem is:
y(x) = c₁ x⁻¹ + c₂ x³ + (1/2) x³ ln x

Example 3: Find the solution of x²y'' - xy' - 3y = 2x³ using Variation of Parameters.

Solution: As noted in the previous examples, the solution of the homogeneous problem has two linearly independent solutions:
y₁(x) = x⁻¹, y₂(x) = x³

Assuming a particular solution of the form:
yₚ(x) = c₁(x) y₁(x) + c₂(x) y₂(x),
we need to solve the system:

c₁'(x) x⁻¹ + c₂'(x) x³ = 0
-c₁'(x) x⁻² + 3c₂'(x) x² = 2x³

From the first equation:
c₁'(x) = -x c₂'(x)

Substituting into the second equation:
c₂'(x) = 1 / 2x, so
c₂(x) = (1/2) ln |x|

Thus, c₁(x) = 1/8 x⁻¹.

The particular solution is:
yₚ(x) = c₁(x) y₁(x) + c₂(x) y₂(x) = (1/8) x⁻¹ + (1/2) x³ ln x

Adding this to the homogeneous solution, we obtain the same solution as in the last example using the Method of Undetermined Coefficients. However, since (1/8) x⁻¹ is a solution of the homogeneous problem, it can be absorbed into the first term, leaving:

y(x) = c₁ x⁻¹ + c₂ x³ + (1/2) x³ ln x